**ASSIGNMENT 8 - PROBLEM 12**

Given an acute triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove:

**PART ONE - CONSTRUCTION:**

In this part, I will just refer to orthocenter file of assignment five.

Let's rename the points so that H is the orthocenter of triangle ABC and D, E, and F are the feet of the perpendiculars from A, B, and C respectfully.

After constructing triangle ABC with the requested conditions, in PART - II below I will prove the first equality and in PART - III, I will prove the second equality.

**PART TWO - PROVE **

Before proving this equality, I would ask my students to play with it using GSP. In this way they can see that what they are trying to prove is really true. Click the figure below to play with the GSP file:

I will prove this equality via **area modeling**. Therefore, first let's see that the **area of ABC is equal to the sum of the areas
of AHB, BHC, and AHC**:

__Here is the proof via area modeling:__

__Step One__: Simplify notation first. Let's define the
area of triangle ABC to be *a*.

__Step Two__: Let's start with the following basic equality

Area Triangle AHB + Area Triangle BHC + Area Triangle AHC = Area Triangle ABC

__Step Three__: Divide each side of the equation by the
Area of Traingle ABC, namely by *a*.

(1/*a*)(Area Triangle AHB + Area Triangle BHC +
Area Triangle AHC ) = 1

__Step Four__: Use distributive property of multiplication
over addition:

(1/*a*)(Area Triangle AHB )+ (1/*a*)(Area Triangle BHC) + (1/*a*)(Area Triangle AHC ) = 1

__Step Five__: Write the areas of each triangle in terms
of theirs bases and heights: AREA = 0.5 x
BASE x HEIGHT

(1/*a*) (0.5) (AB) (FH) + (1/*a*) (0.5) (BC) (DH) + (1/*a*) (0.5) (AC) (EH) = 1

__Step Six__: We are close now to what we want to prove.
It just remains to divide each one of the
three terms on the left hand side by the
appropriate *a* of the Area of Triangle ABC.

__Step Seven__: Area of Triangle ABC, namely *a*, can be expressed in three different ways:

*a* = (0.5) (AB) (FC) , OR , *a* = (0.5) (BC) (AD) , OR , *a* = (0.5) (AC) (BE). The following little
table summarizes this:

FIRST FORM |
SECOND FORM |
THIRD FORM |

a = (0.5) (AB) (FC) |
a = (0.5) (BC) (AD) |
a = (0.5) (AC) (BE) |

__Step Eight__: If we use the first form of *a* with the first term on the left hand side
of the equation in Step Five above, then
it simplifies to AH / FC

__Step Nine__: If we use the second form of *a* with the second term on the left hand side
of the equation in Step Five above, then
it simplifies to DH / AD

__Step Eight__: If we use the third form of *a* with the third term on the left hand side
of the equation in Step Five above, it simplifies
to EH / BE

__Step Ten__: Rewriting the equation of Step Five in
simplified form, we get (FH / FC) + (DH /
AD) + (EH / BE) = 1 and this completes the
proof.

**PART THREE - PROVE**

Before proving this equality, I would ask my students to play with it using GSP. In this way they can see that what they are trying to prove is really true. Click the figure below to play with the GSP file:

I will prove this equality **by using the proof of the first identity** I proved in PART - II above:

__Step One__: DH / AD + EH / BE + FH / CF= 1

__Step Two__: Comparing this expression with what we
want to prove, we see that the denominator
of each term is the same.

__Step Three__: Express DH in terms of AH. Namely, DH =
AD - AH.

__Step Four__: Express EH in terms of BH. Namely, EH =
BE - BH.

__Step Five__: Express FH in terms of CH. Namely, FH =
CF - CH.

__Step Six__: Now plug these in the equation of Step
One:

(AD - AH) / AD + (BE - BH) / BE + (CF - CH) / CF= 1

__Step Seven__: This time, use distributive property of
division over subtraction:

[(AD / AD) - (AH / AD)] + [(BE / BE) - (BH / BE)] + [(CF / CF) - (CH / CF)] = 1

__Step Eight__: Simplify to get

[ 1 - (AH / AD)] + [ 1 - (BH / BE)] + [ 1 - (CH / CF)] = 1

__Step Nine__: Rewrite this as

3 - [ (AH / AD) + (BH / BE) + (CH / CF) ] = 1

__Step Ten__: Solve for the term in the square bracket

[ (AH / AD) + (BH / BE) + (CH / CF) ] = 2

and this completes the proof.

**PART FOUR - WHAT HAPPENS IF ABC IS AN OBTUSE
TRIANGLE?**

What we proved is valid only when the orthocenter H is inside the triangle ABC, in other words, for acute triangles. Indeed, we can see this by clicking the figure below that comprises the two cases together:

**PART FIVE - ORTHOPATH REVIEW:**

In Assignment 4, as an application of parabola, we learned that the trajectory of the orthocenter of a triangle is a parabola when one of the bases is fixed and the vertex opposite to this base is animated along aline that is parallel to this base. In our case, if we fix the base BC while animating the point A along an invisible line parallel to BC, we realize that the orthocenter H traces a parabola. Click the figure below to download this: